Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const double eps = 1e-10;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x > 0) return 1;
  return -1;
}
pair<double, double> lines[maxn];
double get(double x1, double y1, double x2, double y2) {
  if (sgn(x1 - x2) == 0) return x1;
  return x2 - (x2 - x1) * y2 / (y2 - y1);
}
double sum[maxn];
int main() {
  int sy, a, b;
  cin >> sy >> a >> b;
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++)
    scanf("%lf %lf", &lines[i].first, &lines[i].second),
        sum[i] = sum[i - 1] + lines[i].second - lines[i].first;
  sort(lines + 1, lines + 1 + n);
  int q;
  scanf("%d", &q);
  while (q--) {
    double x, y;
    scanf("%lf %lf", &x, &y);
    double left, right;
    left = get(x, y, a, sy);
    right = get(x, y, b, sy);
    int idx1 =
        upper_bound(lines + 1, lines + 1 + n, make_pair(left, left)) - lines;
    idx1--;
    int idx2 =
        upper_bound(lines + 1, lines + 1 + n, make_pair(right, right)) - lines;
    idx2--;
    if (idx1 == idx2) {
      if (sgn(lines[idx1].second - right) >= 0)
        printf("%.15f\n", 1.0 * (b - a) / 1);
      else {
        if (sgn(lines[idx1].second - left) <= 0)
          puts("0");
        else {
          double tmp = lines[idx1].second - left;
          double rate = tmp / (right - left);
          printf("%.15f\n", rate * (b - a));
        }
      }
    } else {
      double tmp = 0;
      if (sgn(lines[idx1].second - left) >= 0)
        tmp += (lines[idx1].second - left);
      if (sgn(lines[idx2].second - right) >= 0)
        tmp += right - lines[idx2].first;
      else
        tmp += lines[idx2].second - lines[idx2].first;
      idx2--;
      tmp += sum[idx2] - sum[idx1];
      double rate = tmp / (right - left);
      printf("%.15f\n", rate * (b - a) / 1);
    }
  }
  return 0;
}